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Chapter 5: Problem 32

An \(85-N\) box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 \(\mathrm{m} / \mathrm{s}\) each second. The push force has a horizontal component of 20 \(\mathrm{N}\) and a vertical component of 25 \(\mathrm{N}\) downward. Calculate the coefficient of kinetic friction between the box and floor.

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.071.

Step by step solution

01

Calculate Gravitational Force

Start by understanding that the gravitational force on the box, due to gravity, is equal to its weight, which is given as 85 N.
02

Determine Normal Force

The normal force is the perpendicular contact force exerted by the floor on the box. We need to consider both the gravitational force (85 N) and the downward component of the push force (25 N) affecting the normal force.The normal force (N) would therefore be: \[ N = 85 + 25 = 110 \text{ N} \]
03

Apply Newton's Second Law

Use Newton's second law in the horizontal direction. The total horizontal force causes the horizontal acceleration. Here, the horizontal force component of 20 N is opposed by the friction force.Since the object is slowing down, the net force is:\[ F_{net} = -f_{kitchen} = m \cdot a \] where \( f_{kitchen} \) is the friction force.
04

Calculate Mass of the Box

To find the mass of the box, use the equation for weight:\[ W = m \cdot g \] Given the weight (W) is 85 N and \(g\) (acceleration due to gravity) is 9.8 m/s虏, mass (m) is:\[ m = \frac{85}{9.8} \approx 8.67 \text{ kg} \]
05

Determine Deceleration

The box is slowing down at a rate of 0.90 m/s虏, meaning it has a negative acceleration (deceleration).\[ a = -0.90 \text{ m/s}^2 \]
06

Calculate Frictional Force

Using Newton鈥檚 second law, find the frictional force:\[ f_{kitchen} = m \cdot a \ f_{kitchen} = 8.67 \times (-0.90) \approx -7.80 \text{ N}\]
07

Calculate Coefficient of Kinetic Friction

The friction force is also given by:\[ f_{kitchen} = \mu_k \cdot N \] Rearrange to solve for \( \mu_k \):\[ \mu_k = \frac{|f_{kitchen}|}{N} = \frac{7.80}{110} \approx 0.071 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law forms the foundation of understanding motion. It states that the acceleration of an object is produced when a net force acts on it. The law is expressed by the equation:\[ F_{net} = m imes a \]where:
  • \( F_{net} \) is the net force acting on the object
  • \( m \) is the mass of the object
  • \( a \) is the acceleration
In this exercise, Newton's Second Law helps us understand how the frictional force interacts with the motion of the box. The box is subjected to a horizontal force, which causes an acceleration. However, because it's slowing down, the net force is negative, showing that the friction force is greater than the horizontal force pushing the box.
To calculate the net horizontal force, we consider the forces in the horizontal direction. The opposing frictional force leads to a deceleration of the box as it moves, with the following relationship:\[ F_{net} = m imes (-0.90) = -f_{kitchen} \]where \(-0.90\) m/s虏 is the deceleration of the box. This relationship not only shows the importance of mastering force dynamics but also highlights how essential Newton's Second Law is in analyzing motion.
Frictional Force
Frictional force opposes the motion between two surfaces in contact. In this context, kinetic friction is the type of friction acting on the moving box. It plays a crucial role in our problem since it counters the horizontal force pushing the box, causing it to slow down.
The frictional force can be calculated using the equation:\[ f_{kitchen} = ext{frictional force} = ext{mass} \times ext{deceleration} \]In our example:
  • The mass of the box is approximately 8.67 kg.
  • The deceleration is \(-0.90\) m/s虏.
By plugging these values into the formula, we get:\[ f_{kitchen} = 8.67 \times (-0.90) \approx -7.80 \text{ N} \]This result shows us the magnitude of the frictional force acting against the push force. Since friction works in opposition to applied forces, understanding it is essential for solving any problem involving motion and resistance, such as determining the coefficient of kinetic friction.
Normal Force
The normal force is the support force exerted by a surface, perpendicular to the object resting on it. For this specific problem, the normal force is critical because it is used to calculate the frictional force that resists the motion of the box.
To find the normal force, we must take into account both the weight of the box and any additional vertical force components. Here, the force components are:
  • Weight of the box: 85 N
  • Downward vertical force from the push: 25 N
The sum of these forces gives the normal force \( N \):\[ N = 85 + 25 = 110 \text{ N} \]This normal force serves as the baseline for calculating the kinetic friction force since friction is proportional to how forcefully surfaces are pressed together. Understanding the normal force is crucial for calculating the coefficient of kinetic friction, represented as \( \mu_k \), which influences how easily objects slide over each other.

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Most popular questions from this chapter

Block \(B\) , with mass 5.00 \(\mathrm{kg}\) , rests on block \(A\) , with mass 8.00 \(\mathrm{kg}\) , which in turn is on a horizontal tabletop (Fig. 5.72 ). There is no friction between block \(A\) and the tabletop, but the coefficient of static friction between block \(A\) and block \(B\) is \(0.750 .\) A light string attached to block \(A\) passes over a frictionless, massless pulley, and block \(C\) is suspended from the other end of the string. What is the largest mass that block \(C\) can have so that blocks \(A\) and \(B\) still slide together when the system is released from rest? figure can't copy

You observe a 1350 -kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to thesquare of its speed). You take the following data during a time interval of \(25 \mathrm{s} :\) When its speed is 32 \(\mathrm{m} / \mathrm{s}\) , the car slows down at a rate of \(-0.42 \mathrm{m} / \mathrm{s}^{2},\) and when its speed is decreased to \(24 \mathrm{m} / \mathrm{s},\) it slows down at \(-0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the coefficient of rolling friction and the air drag constant \(D .(b)\) At what constant speed will this car move down an incline that makes a \(2.2^{\circ}\) angle with the horizontal? (c) How is the constant speed for an incline of angle \(\beta\) related to the terminal speed of this sports car if the car drops off a high cliff? Assume that in both cases the air resistance force is proportional to the square of the speed, and the air drag constant is the same.

Falling Baseball. You drop a bascball from the roof of a tall building. As the ball falls, the air exerts a drag force proportional to the square of the ball's speed \(\left(f=D v^{2}\right) .\) (a) In a diagram, show the direction of motion and indicate, with the aid of vectors, all the forces acting on the ball. (b) Apply Newton's second law and infer from the resulting equation the general properties of the motion. (c) Show that the ball acquires a terminal speed that is as given in Eq. (5.13). (d) Derive the equation for the speed at any time. (Note: $$ \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{a} \operatorname{arctanh}\left(\frac{x}{a}\right) $$ where $$ \tanh (x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{e^{2 x}-1}{e^{2 x}+1} $$ defines the hyperbolic tangent.)

A \(125-\mathrm{kg}\) (including all the contents) rocket has an engine that produces a constant vertical force (the thrust) of 1720 \(\mathrm{N}\) . Inside this rocket, a \(15.5-\mathrm{N}\) electrical power supply rests on the floor. (a) Find the acceleration of the rocket, (b) When it has reached an altitude of 120 \(\mathrm{m}\) , how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)

The Monkey and Bananas Problem. A \(20-k g\) monkey has a firm hold on a light rope that passes over a frictionless pulley and is attached to a \(20-\mathrm{kg}\) bunch of bananas (Fig. 5.77\()\) . The monkey looks up, sees the bananas, and starts to climb the rope to get them. (a) As the monkey climbs, do the bananas move up, down, or remain at rest? (b) As the monkey climbs, does the distance between the monkey and the bananas decrease, increase, or remain constant? (c) The monkey releases her hold on the rope. What happens to the distance between the monkey and the bananas while she is falling?(d) Before reaching the ground, the monkey grabs the rope to stop her fall. What do the bananas do? figure can't copy
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